Proof Involving Rational and Irrational Numbers

Date: 09/13/2004 at 10:23:43
From: John
Subject: Proving existence of irrational a, b, where a^b is rational
The question I'm stuck on is:
Prove the existence of 2 irrational numbers, a and b, where a^b is
rational.
Thanks for any help!

Date: 09/13/2004 at 10:50:35
From: Doctor Luis
Subject: Re: Proving existence of irrational a, b, where a^b is rational
Hi John,
Good question! This has been considered before, and there's a very
interesting (if not famous) proof of this result. The proof is
interesting in the sense that it is non-constructive, i.e. we never
actually find specific irrational values a,b that are both decidedly
irrational. Nonetheless, we can still prove the theorem.
We start by noting that sqrt(2) is irrational and that 2 is rational.
Next, we define the number x = sqrt(2)^sqrt(2).
By logic (law of excluded middle), we can confidently say that our
number x is either rational or it is irrational.
So, we consider each of the two possibilities:
1) Say x is rational. Then we are done! We have found
two values a = sqrt(2), b = sqrt(2) such that a^b is rational.
The theorem is proven, at least for this possibility.
2) Say x is irrational. Then we are done, too! Take
a = x = sqrt(2)^sqrt(2), and b = sqrt(2). Then a^b = 2
is rational. The theorem is proven for the second possibility.
Therefore the theorem is true because we have exhausted the two
possibilities. The interesting part is that it doesn't matter whether
x is rational or irrational! Either way, the theorem is proven.
Incidentally, you can actually PROVE that x = sqrt(2)^sqrt(2) is
irrational by using a result known as Gelfond's theorem. You can read
more about it if you follow these URLs:
Gelfond's Theorem
http://mathworld.wolfram.com/GelfondsTheorem.html
Gelfond-Schneider Constant: 2^sqrt(2)
http://mathworld.wolfram.com/Gelfond-SchneiderConstant.html
However, our little proof is quite elegant, albeit indirect.
Let us know if you have any other thought-provoking questions.
- Doctor Luis, The Math Forum
http://mathforum.org/dr.math/